Logic Coverage from Source Code Moonzoo Kim School of Computing KAIST The original slides are taken from Chap. 8 of Intro. to SW Testing 2nd ed by Ammann and Offutt

Logic Expressions from Source Predicates are derived from decision statements in programs In programs, most predicates have less than four clauses Wise programmers actively strive to keep predicates simple When a predicate only has one clause, COC, ACC, ICC, and CC all collapse to predicate coverage (PC) Applying logic criteria to program source is hard because of reachability and controllability: Reachability : Before applying the criteria on a predicate at a particular statement, we have to get to that statement Controllability : We have to find input values that indirectly assign values to the variables in the predicates Variables in the predicates that are not inputs to the program are called internal variables These issues are illustrated through the triangle example in the following slides …

30 private static int Triang (int s1, int s2, int s3) 31 { 32 int result; 33 34 // result is output from the routine: 35 // result = 1 if triangle is scalene 36 // result = 2 if triangle is isosceles 37 // result = 3 if triangle is equilateral 38 // result = 4 if not a triangle 39 40 // After a quick confirmation that it’s a legal 41 // triangle, detect any sides of equal length 42 if (s1 <= 0 || s2 <= 0 || s3 <= 0) 43 { 44 result = 4; 45 return (result); 46 } 47 48 result = 0; 49 if (s1 == s2) 50 result = result + 1; 51 if (s1 == s3) 52 result = result + 2; 53 if (s2 == s3) 54 result = result + 3; 55 if (result == 0) 56 { // Confirm it’s a legal triangle before declaring 57 // it to be scalene 59 if (s1+s2<=s3||s2+s3 <= s1 60 || s1+s3 <= s2) 61 result = 4; 62 else 63 result = 1; 64 return (result); 65 } 67 /* Confirm it’s a legal triangle before declaring 68 it to be isosceles or equilateral */ 69 70 if (result > 3) 71 result = 3; 72 else if (result == 1 && s1+s2 > s3) 73 result = 2; 74 else if (result == 2 && s1+s3 > s2) 75 result = 2; 76 else if (result == 3 && s2+s3 > s1) 77 result = 2; 78 else 79 result = 4; 80 return (result); 81 } // end Triang

Ten Triangle Predicates 42: (s1 <= 0 || s2 <= 0 || s3 <= 0) 49: (s1 == s2) 51: (s1 == s3) 53: (s2 == s3) 55: (result == 0) 59: (s1+s2 <= s3 || s2+s3 <= s1 || s1+s3 <= s2) 70: (result > 3) 72: (result == 1 && s1+s2 > s3) 74: (result == 2 && s1+s3 > s2) 76: (result == 3 && s2+s3 > s1)

Reachability for Triang Predicates 42: True 49: P1 = s1>0 && s2>0 && s3>0 51: P1 53: P1 55: P1 59: P1 && result = 0 70: P1 && result != 0 72: P1 && result != 0 && result <= 3 74: P1 && result != 0 && result <= 3 && (result !=1 || s1+s2<=s3) 76: P1 && result != 0 && result <= 3 && (result !=1 || s1+s2<=s3) && (result !=2 || s1+s3<=s2) Need to solve for the internal variable result

Solving for Internal Variable result At line 55, result has a value in the range (0 .. 6) result = 0 s1!=s2 && s1!=s3 && s2!=s3 1 s1=s2 && s1!=s3 && s2!=s3 2 s1!=s2 && s1=s3 && s2!=s3 3 s1!=s2 && s1!=s3 && s2=s3 4 s1=s2 && s1!=s3 && s2=s3 5 s1!=s2 && s1=s3 && s2=s3 6 s1=s2 && s1=s3 && s2=s3 Contradiction Contradiction

Reachability for Triang Predicates (solved for result – reduced) 42: True 49: P1 = s1>0 && s2>0 && s3>0 51: P1 53: P1 55: P1 59: P1 && s1 != s2 && s2 != s3 && s2 != s3 (result = 0) 70: P1 && P2 = (s1=s2 || s1=s3 || s2=s3) (result != 0) 72: P1 && P2 && P3 = (s1!=s2 || s1!=s3 || s2!=s3) (result <= 3) 74: P1 && P2 && P3 && (s1 != s2 || s1+s2<=s3) 76: P1 && P2 && P3 && (s1 != s2 || s1+s2<=s3) && (s1 != s3 || s1+s3<=s2) Looks complicated, but a lot of redundancy

Predicate Coverage T F s1 s2 s3 s1 s2 s3 These values are “don’t care”, needed to complete the test. p42: (s1 <= 0 || s2 <= 0 || s3 <= 0) p49: (s1 == s2) p51: (s1 == s3) p53: (s2 == s3) p55: (result == 0) p59: (s1+s2 <= s3 || s2+s3 <= s1 || s1+s3 <= s2) p70: (result > 3) p72: (result == 1 && s1+s2 > s3) p74: (result == 2 && s1+s3 > s2) p76: (result == 3 && s2+s3 > s1) T F s1 s2 s3 s1 s2 s3 0 0 0 1 1 1 1 1 1 1 2 2 1 1 1 2 1 2 1 2 3 1 1 1 1 2 3 2 3 4 1 1 1 2 2 3 2 2 3 2 2 4 2 3 2 2 4 2 3 2 2 4 2 2

Clause Coverage T F p42: (s1 <= 0) 0 1 1 4 1 1 1 3 (s2 <= 0 ) p59: (s1+s2 <= s3 ) (s2+s3 <= s1) (s1+s3 <= s2) p72: (result == 1) (s1+s2 > s3) p74: (result == 2) (s1+s3 > s2) p76: (result == 3) (s2+s3 > s1) T F S1 s2 s3 EO s1 s2 s3 EO 0 1 1 4 1 1 1 3 1 0 1 4 1 1 1 3 1 1 0 4 1 1 1 3 2 3 6 4 2 3 4 1 6 2 3 4 2 3 4 1 2 6 3 4 2 3 4 1 2 2 3 2 2 3 2 2 2 2 3 2 2 2 5 4 2 3 2 2 3 2 2 2 2 3 2 2 2 5 2 4 3 2 2 2 1 2 1 4 3 2 2 2 5 2 2 4

CACC Coverage (also RACC) c1 c2 c3 P s1 s2 s3 EO p42: (s1 <= 0 || s2 <= 0 || s3 <= 0) p59: (s1+s2 <= s3 || s2+s3 <= s1 || s1+s3 <= s2) p72: (result == 1 && s1+s2 > s3) p74: (result == 2 && s1+s3 > s2) p76: (result == 3 && s2+s3 > s1) T f f t 0 1 1 4 F F F f 1 1 1 3 f T f t 1 0 1 4 f f T t 1 1 0 4 T f f t 2 3 6 4 F F F f 2 3 4 1 f T f t 6 2 3 4 f f T t 2 6 3 4 T T t 2 2 3 2 F t f 2 3 3 2 t F f 2 2 5 4 T T t 2 3 2 2 t F f 2 5 2 4 T T t 3 2 2 2 F t f 1 2 2 4 t F f 5 2 2 4 s1=s2 && s1!=s3 && s2!=s3 s1!=s2 && s1=s3 && s2!=s3 s1!=s2 && s1!=s3 && s2=s3 Same answers to RACC

Program Transformation Issues if ((a && b) || c) { s1; } else { s2; if (a) { if (b) s1; else { if (c) /* c1 */ else s2; } if (c) /* c2 */ Transform (1)? Transform (2)? d = a && b; e = d || c; if (e) { s1; } else { s2; Conjecture: transform a program to make generating required test cases easy. But now working! 11 11

Problems with Transformed Programs (1/2) Maintenance is certainly harder with Transform (1) Not recommended! Coverage on Transform (1) PC on the transform does not imply CACC on original A test suit to satisfy PC on the transform (1): a:any element of {1,2,3,4}x{5,6,7,8} b:any element of {1,2}x{3,4} c1:{(3,4)} c2:any element of {5,7}x{6,8} ex. {1,3,4,5,8} CACC on the original does not imply PC on transform Ex. {(2,6),(2,4),(3,4)} does not satisfy PC on the transform due to c2 a b c (ab)c CACC PC(1) 1 T O 2 F 3 4 5 6 7 8 (ab)c a as major clause: pa: b ¬c TR={(2,6)} b as major clause: pb: a ¬c TR={(2,4)} c as major clause: pc: ¬(ab) TR= any element of {3,5,7}x{4,6,8}

Problems with Transformed Programs (2/2) Coverage on Transform (2) Structure used by logic criteria is “lost” Hence CACC on the transform 2 only requires 3 tests Therefore, it may not be meaningful to transform a program to increase coverage a b d c (ab)c CACC PC(1) CACC(2) 1 T O 2 F 3 4 5 6 7 8 d || c d as major clause: pd: ¬c TR={(2,4),(2,6),(2,8)} c as major clause: pc: ¬d TR={3,5,7}x{4,6,8}

Summary : Logic Coverage for Source Code Predicates appear in decision statements if, while, for, etc. Most predicates have less than four clauses But some applications have predicates with many clauses The hard part of applying logic criteria to source is resolving the internal variables Non-local variables (class, global, etc.) are also input variables if they are used If an input variable is changed within a method, it is treated as an internal variable thereafter To maximize effect of logic coverage criteria: Avoid transformations that hide predicate structure 14 14

Restricted Active Clause Coverage s1 s2 s3 EO p42: (s1 <= 0 || s2 <= 0 || s3 <= 0) p59: (s1+s2 <= s3 || s2+s3 <= s1 || s1+s3 <= s2) p72: (result == 1 && s1+s2 > s3) p74: (result == 2 && s1+s3 > s2) p76: (result == 3 && s2+s3 > s1) T f f t 0 1 1 4 F F F f 1 1 1 3 f T f t 1 0 1 4 f f T t 1 1 0 4 T f f t 2 3 6 4 F f f f 2 3 4 1 f T f t 6 2 3 4 f f T t 2 6 3 4 T t t 2 2 3 2 F t f 2 3 3 2 t F f 2 2 5 4 T t t 2 3 2 2 t F f 2 5 2 4 T t t 3 2 2 2 F t f 1 2 2 4 t F f 5 2 2 4 s1=s2 && s1!=s3 && s2!=s3 s1!=s2 && s1=s3 && s2!=s3 s1!=s2 && s1!=s3 && s2=s3